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murzikaleks [220]
2 years ago
13

How many four-digit personal identification numbers (PIN) are possible if the PIN cannot begin or end with a 0 and the PIN must

be an even number? a. 3,600 b. 4,000 c. 4,500 d. 8,100
Mathematics
2 answers:
Anna35 [415]2 years ago
8 0

Consider 4 positions of four-digit PIN:

1. PIN cannot begin with 0, then it can begin with 9 remaining digits (1, 2, 3, 4, 5, 6, 7, 8 and 9) - 9 posibilities;

2. for the second and third position yuo can select any of 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) - 10·10=100 posibilities;

3. the PIN must be an even number and cannot be 0, then it ends with any of 4 even digits (2, 4, 6 and 8) - 4 posibilities.

Use the product rule to count the total number of posibilities:

9·100·4=3600.

Answer: correct choice is A.

Masteriza [31]2 years ago
4 0
There would be 4500 combonations
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